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$ f(x) = 6x^7\sin^2(x^{1000}) e^{x^2} $

Find $ f^{(2013)}(0) $

A math forum friend suggest me to use big O symbol, however have no idea what that is, so how does that helping?

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11  
Where did you find this problem? 2013 is an odd year to choose... :) –  anorton Dec 30 '12 at 20:58
5  
$\sin^2(x^{1000}) = x^{2000}+O(x^{4000})$ reduces the complexity of the computation considerably... –  WimC Dec 30 '12 at 21:04

2 Answers 2

up vote 42 down vote accepted

Note that,

$$ 6\,x^{7} \sin\left(x^{1000}\right)\sin\left(x^{1000}\right)e^{x^2} $$

$$ = 6\,x^{7} \left( x^{1000}-\frac{x^{3000}}{3!}+\dots \right)\left( x^{1000}-\frac{x^{3000}}{3!}+\dots \right)\left(1+\frac{x^2}{1!}+\frac{x^4}{4!}+\dots\right) $$

$$ = 6x^7x^{2000}\left( 1-\frac{x^{2000}}{3!} +\dots\right)^2\left(1+\frac{x^2}{1!}+\frac{x^4}{2!}+\dots\right) $$

$$ = 6x^{2007}\left(1+\frac{x^2}{1!}+\frac{x^4}{2!}+\frac{x^6}{3!}+\dots\right)\left( 1-\frac{x^{2000}}{3!} +\dots\right)^2 $$

Now, it is clear that the coefficient of $x^{2013}$ is $1$, which implies that

$$ \frac{f^{(2013)}(0)}{(2013)!} = 1 \implies f^{(2013)}(0)=(2013)!. $$

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Nice job there (+1) –  Chris's sis Dec 31 '12 at 20:56
    
@Chris'ssister: Thank you. –  Mhenni Benghorbal Dec 31 '12 at 23:08

Hint:

Consider the Taylor expansion of $f$

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1  
Short and sweet...+1 :-) –  amWhy Dec 30 '12 at 21:14

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